Alphabeta Math
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

7 results · all verified · 7 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs; all 7 also cleared it.

Rees Modules Artin Rees and Hilbert Samuel Theory — Examples

1 · Prerequisites

2 · Summary

These examples make the abstract graded and Hilbert-Samuel constructions computable. They show how Hilbert series are counted in a polynomial ring and a homogeneous quotient, how the associated graded ring records a tangent cone, how Artin-Rees and Krull intersection look in explicit one-variable cases, and how Hilbert-Samuel polynomials and multiplicities appear in DVR, cusp, and finite-length settings.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The polynomial ring and a homogeneous quotient have the expected Hilbert series and Hilbert polynomial

Example

Let k be a field and give k[x,y] the standard grading. Then

HSk[x,y](t)=1(1t)2,

because the degree-n piece has basis

xn,xn1y,,xyn1,yn

and therefore dimension n+1.

For the homogeneous quotient

A:=k[x,y]/(y2),

the degree-0 piece has dimension 1 and each degree-n1 piece has basis xn,xn1y. Hence

HSA(t)=1+2t+2t2+=1+t1t,

so the Hilbert polynomial of A is the constant polynomial 2.

Facts & Assumptions

Given: A field k, the standard grading on k[x,y], and the quotient A=k[x,y]/(y2).

[L1]

Finite graded modules over standard graded algebras have rational Hilbert series and eventual polynomial growth (A finite graded module over a standard graded algebra has rational Hilbert series and eventual polynomial growth).

Verification

technique · direct
1.1

In k[x,y], the degree-n monomials are exactly xniyi for 0in, so the degree-n piece has dimension n+1. Therefore HSk[x,y](t)=n0(n+1)tn=1(1t)2.

givenalgebra
1.2

In the quotient by (y2), every monomial containing y2 vanishes. So for n1 the degree-n piece is spanned by xn and xn1y, and these two classes are linearly independent. Hence HSA(t)=1+n12tn=1+t1t.

givenalgebra
1.3

The eventual coefficient sequence of HSA(t) is constant equal to 2, so the Hilbert polynomial is 2; this matches the general rationality promised by [L1].

L1
2.1

Thus both the polynomial ring and this homogeneous quotient realize the expected Hilbert series and Hilbert polynomial.

algebra
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The associated graded ring of a regular local ring and of a cusp local ring can be computed explicitly

Example

Let

R1:=k[x,y](x,y)

with maximal ideal m=(x,y). Then

grm(R1)k[X,Y],

because mn/mn+1 has basis given by degree-n monomials in the initial classes of x and y.

For the cusp local ring

R2:=k[x,y](x,y)/(y2x3),

the initial form of the relation has degree 2, so

grm(R2)k[X,Y]/(Y2).

Facts & Assumptions

Given: A field k, the local rings R1 and R2 above, and the maximal-ideal filtrations.

[L1]

The associated graded ring is

grm(R)=n0mn/mn+1

(The associated graded ring and associated graded module of an ideal-adic filtration).

Verification

technique · direct
1.1

In R1, the classes of x and y in m/m2 generate every graded piece: the images of the degree-n monomials xniyi form a basis of mn/mn+1. Therefore the map k[X,Y]grm(R1) sending X,Y to the initial classes of x,y is a graded isomorphism.

L1givenalgebra
1.2

In R2, the relation y2x3 lies in m2 and its lowest-degree term is y2. Hence the only initial relation in degree 2 is Y2=0. As in the remaining monomials Xn and Xn1Y survive and span the graded pieces, so grm(R2)k[X,Y]/(Y2).

L1algebra
2.1

These explicit computations exhibit the regular local and cusp cases.

algebra
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

An explicit Artin-Rees number can be computed for a submodule inside a finite module

Example

Take R=k[x], I=(x), M=R, and N=(xm) for a fixed integer m0. Then for every nm,

InMN=(xn)(xm)=(xn)=Inm(ImMN).

So in this case one may take the Artin-Rees number to be c=m.

Facts & Assumptions

Given: A field k, an integer m0, the ring R=k[x], the ideal I=(x), the module M=R, and the submodule N=(xm).

[L1]

Artin-Rees gives some constant c with

InMN=Inc(IcMN)

for all nc (Artin-Rees controls intersections of submodules with high ideal powers).

Verification

technique · direct
1.1

Here InM=(xn) and N=(xm). If nm, then (xn)(xm), so InMN=(xn).

givenalgebra
1.2

Also ImMN=(xm), and therefore for nm, Inm(ImMN)=(xnm)(xm)=(xn)=InMN.

algebra
2.1

So c=m works, exhibiting an explicit Artin-Rees bound compatible with the abstract existence statement [L1].

L1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

In a Noetherian local domain, the intersection of the powers of the maximal ideal is zero

Example

Assume the Axiom of Choice.

Let (R,m) be a Noetherian local domain. Then

n0mn=0.

Facts & Assumptions

Given: The Axiom of Choice and a Noetherian local domain (R,m).

[L1]

The Krull intersection theorem says that for a finite module M,

n0mnM=0

when mJ(R) (The Krull intersection is the (1a)-torsion submodule, and it vanishes in the Jacobson-radical case).

Verification

technique · direct
1.1

View R as a finite module over itself. Since (R,m) is local, its maximal ideal lies in the Jacobson radical. Therefore [L1] applies with M=R and gives n0mn=0.

L1given
2.1

This is the promised local-domain instance of Krull intersection.

algebra
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

A DVR has Hilbert-Samuel polynomial n+1 and multiplicity one

Example

Let (V,(π)) be a discrete valuation ring. Then

χ(π),V(n)=V(V/(πn+1))=n+1

for every n0. Hence the Hilbert-Samuel polynomial is exactly n+1, and the Hilbert-Samuel multiplicity is

e(π)(V)=1.

Facts & Assumptions

Given: A discrete valuation ring V with maximal ideal (π).

[L1]

The quotient V/(πn+1) has length n+1 for every n0 (Length and valuation in a DVR).

[L2]

Verification

technique · direct
1.1

By [L1], the Hilbert-Samuel function is χ(π),V(n)=V(V/(πn+1))=n+1 for every n0.

L1given
1.2

Therefore the eventual polynomial is already exactly P(n)=n+1. Its degree is 1 and its leading coefficient is 1, so [L2] gives e(π)(V)=1!1=1.

L2
2.1

This computes both the Hilbert-Samuel polynomial and the multiplicity of a DVR.

algebra
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The Hilbert-Samuel multiplicity of a plane-curve singularity is read from its associated graded ring

Example

Let

R:=k[x,y](x,y)/(y2x3)

with maximal ideal m=(x,y)/(y2x3). Then

grm(R)k[X,Y]/(Y2),

so the homogeneous piece of degree n1 has basis Xn,Xn1Y and dimension 2. Consequently

χm,R(n)=1+j=1n2=2n+1

for n1, and therefore

em(R)=2.

Facts & Assumptions

Given: A field k, the cusp local ring R above, and its maximal ideal m.

[L1]

The associated graded ring packages the quotients mn/mn+1, and the Hilbert-Samuel function is their cumulative length (The associated graded ring and associated graded module of an ideal-adic filtration, The Hilbert-Samuel function agrees eventually with a rational polynomial in binomial form).

[L2]

Hilbert-Samuel multiplicity is the leading coefficient scaled by the factorial (Hilbert-Samuel multiplicity as the factorial-scaled leading coefficient).

Verification

technique · direct
1.1

The initial form of y2x3 has degree 2, namely Y2, so grm(R)k[X,Y]/(Y2). Thus the degree-0 piece has dimension 1, and every degree-n1 piece has basis Xn,Xn1Y.

L1givenalgebra
1.2

Therefore R(m0/m)=1 and R(mn/mn+1)=2(n1). Summing these lengths as in [L1] gives χm,R(n)=2n+1 for n1.

L1algebra
1.3

The eventual polynomial is 2n+1, so its degree is 1 and the leading coefficient is 2. Hence [L2] gives em(R)=2.

L2
2.1

Thus the multiplicity of the cusp is read directly from its tangent-cone graded ring.

algebra
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

In dimension zero the Hilbert-Samuel polynomial is constant and equals the module length

Example

Let (R,m) be a zero-dimensional Noetherian local ring and let M be a finite R-module. Then some power of m annihilates M, so for all sufficiently large n,

M/mn+1MM.

Hence the Hilbert-Samuel polynomial is the constant polynomial

Pm,M(n)=R(M).

Facts & Assumptions

Given: A zero-dimensional Noetherian local ring (R,m) and a finite R-module M.

[L1]

The length R(M) is defined for finite-length modules (Composition series and length of a module).

Verification

technique · direct
1.1

In a zero-dimensional Noetherian local ring, the maximal ideal is nilpotent on every finite module, so there is N with mNM=0. Hence for every nN1, M/mn+1M=M.

givenalgebra
1.2

Therefore the Hilbert-Samuel function is eventually constant equal to R(M), which is defined by [L1]. So the eventual polynomial provided by [L2] is the constant polynomial R(M).

L1L2
2.1

This is exactly the zero-dimensional case.

algebra

Sources